Spacecraft depressurization Let's imagine a spacecraft has an internal pressure $P_0$ and a hole is pierced on its hull. I am asked to prove that the internal pressure decreases exponentially. This is trivial if we can prove that
$$ \tag{1} \frac{dP}{dt}=-kP(t) $$
but I have no idea how to derive that formally.
More in general, how can you prove that a gas moves opposite its gradient of pressure, at a rate which depends on the pressure itself?
EDIT
The only possibile solution that comes to my mind that doesn't make use of advanced concepts is the following:
The number of particles per unit time (and thus the number of moles) passing through the cross sectional area of the hole is proportional to the internal pressure:
$$ \tag{2} \frac{dn}{dt}=-kP(t)$$
Now differentiating the perfect gas law
$$ dP \cdot V = dn \cdot R T $$
I have assumed that the temperature doesn't change because this is an adiabatic process and no work is done by the gas. Now substituting:
$$ dP \cdot V = -kP(t)dt\cdot RT$$ 
$$ \frac{dP}{dt}=-\frac{kRT}{V}P(t) $$.
However, I'm still not sure how to justify (2), and it seems to me that this calculations are just a way of "postponing" the problem.
 A: If the hole is small enough, this is a question about "choked" flow, which is a special case of flow in a duct of variable area. Are any of those words familiar?
EDIT I have tried to find as simple a treatment of duct flow as I could online. Best that I found is
web.mit.edu/16.unified/www/SPRING/fluids/Spring2008/LectureNotes/f19.pdf
To apply this to your problem, note that the flow can only reach a local Mach number of 1.0 at a "throat", a place where the duct area is locally minimum. Treat the air inside the spaceship as a uniform reservoir of stationary air, and the hole as a duct through which it leaves. This "duct" has a minimum area, the area of the hole itself. Imagine that the pressure in the spacecraft is actually very low, so that air blows out gently, but now imagine increasing that pressure. The emerging air gets faster and faster until the Mach number at the hole is 1.0. This is called the choking condition because further increase in the internal pressure will not increase the flow rate. The hole is leaking at the fastest rate possible, and will continue in this choked condition until enough air has leaked out to lower the internal pressure sufficiently. I think this gives you enough to work on the question, but you will have to figure out how to use the information. 
As I said, I am very surprised that this question would be set in this context, but if you are interested enough then give it a shot. You will soon come to realize that all those scenes in airplane movies where someone blows a hole in the cabin wall are ludicrously incompatible with physics.
If you do get both interested and stuck, then give me a shout.
EDIT 2
Let m be the mass of air with density $\rho_s$remaining in the spacecraft, which has volume V_s. $$\frac{dm}{dt}=V_s\frac{d\rho_s}{dt}=\rho_hU_hA_h\rightarrow\frac{d\rho_s}{dt}=\frac{A_h}{V_s}(\rho_hU_h)$$
To move forward, we need to say something about the product $\rho_hU_h$ which concerns properties of the air passing through the hole. This can be found from dimensional analysis, but only if you are familiar with the idea of choked flow. This says if the hole is small (air would gush violently out of a large one) and the pressure ratio across the hole is large (It will be effectively infinite because we have a near-vacuum outside) then conditions at the hole are independent of anything except conditions in the spacecraft.
You are not correct in assuming constant temperature and I am not sure how you relate this to no work being done. The proper assumption is constant entropy because no irreversible processes are involved. Now you have all the bits. 
I was sort of correct that there is what looks like a simple solution but it is rather sophisticated, and not likely to be apparent to anybody with no background in gas dynamics.
